仅打印drupal field_view_field值

问题描述

我正在使用下面的代码将节点的字段打印到特定的区域，它的效果非常好。但是，这是一个我只想打印你没有标签的字段的值的实例。似乎应该很容易，但我有一点麻烦。我很感激任何帮助，因为我是新来的dr​​upal。感谢

I'm using the code below to print the out the field of nodes to specific areas and it works great. But theres an instance where I just want to print the value you of field without the label. Seems as it should be pretty easy but I'm having a bit of trouble. I'd appreciate any help as i'm pretty new to drupal. Thanks

<?php
  print drupal_render(field_view_field('node', $node, 'field_description')); ?>

推荐答案

field_view_value（） code>需要一个 $ display 参数，您可以使用它来隐藏标签：

field_view_value() takes a $display argument that you can use to hide the label:

$display = array('label' => 'hidden');
$view = field_view_field('node', $node, 'field_description', $display);
print drupal_render($view);

如果您只想提取字段的原始值，您可以使用 field_get_items（） / a>代替：

If you just want to extract the raw value of the field you can use field_get_items() instead:

$items = field_get_items('node', $node, 'field_description');
$first_item = array_shift($items);
$description = $first_item['value'];

列名称（ $ first_item ['whatever'] ）将取决于您使用的字段类型。对于文本字段，它将是值。在输出之前，请记住使用 check_plain（）对输入进行消毒，因为Drupal的约定是存储原始输入数据并在输出时对其进行清理。

The column name ($first_item['whatever']) will depend on the type of field you're using. For text fields it will be value. Remember to sanitise the input with check_plain() before you output it as Drupal's convention is to store the raw input data and sanitise it upon output.

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